Question

Помогите решить 29 номер

Answer 1

#29
1.  z=arctg(x+y)(1-xy)   z'x=1/(1+((x+y)(1-xy))²*[(x+y)(1-xy)]'x
[(x+y)(1-xy)]'x=(u'v+v'u)={u=x+y  u'=1   v=1-xy  v'=-y}=(1-xy)+(-y*x-y²)
z'x=[(1-xy)+(-y*x-y²)]/(1+((x+y)(1-xy))²

z'y=1/(1+((x+y)(1-xy))²*[(x+y)(1-xy)]'y
[(x+y)(1-xy)]'y={u=x+y  u'=1   v=1-xy  v'=-x}=[(1-xy)+(-y*x-x²)]
z'y=[(1-xy)+(-y*x-x²)]/(1+((x+y)(1-xy))²

2. z=(ln(x²-y²))²+2arccos√y/√(x-y)
    z'x=2ln(x²-y²)*1/(x²-y²)*2x-2/√(1-y/(x-y))*[√y/√(x-y)]'x
[√y/√(x-y)]'x=[y*(x-y)^(-1/2)]'x=-1/2*(x-y)^(-3/2)
z'x=2ln(x²-y²)*1/(x²-y²)*2x-2/√(1-y/(x-y))*[-1/2*(x-y)^(-3/2)]
z'y=-4ln(x²-y²)*1/(x²-y²)*2y - 2/√(1-y/(x-y))[√y/√(x-y)]'y
[√y/√(x-y)]'y={u=√y   u'=1/2√y   v=√(x-y)  v'y=-1/2√(x-y) }=
=1/(x-y)*[√(x-y)/2√y+√y/2√(x-y)]
z'y=-4ln(x²-y²)*1/(x²-y²)*2y - 2/√(1-y/(x-y))[√(x-y)/2√y+√y/2√(x-y)]